Analysis of Advanced Networks - Assignment Example

Analysis of Advanced Networks Question 1 Three stations, A, B and C, are attached through a 25m, Cat 5, 100Base-TX, Ethernet cable, as shown in figure 2.1. The stations exchange frames of 10000bits on average (including all headers and preambles). The signals propagate at a speed of 200m/sec (1000bits delay for each 200m). The time slot, and the acknowledgement length are 512 and 200 bits respectively. The latencies of the hub and the switch, the 48 bits jamming signal and the 96 bit processing time of the acknowledgment, are both ignored. Figure 2.1: Four stations connected to a hub and a switch a) Briefly, explain the role of each of the following fields shown in figure 2.2, preamble, destination, length, pad, and CRC. Solution: Preamble Preamble is basically a signalling that passes pulses on the network depending on the technology used. Communication between two or multiple nodes with synchronized transmission of data is called a preamble signal. Time factor and tempo emphasizes on the transfer of data between the source and destination is crucial. Field Delimiter Field delimiter specifies boundaries in a data transmission. Example should be when we are writing multiple attributes of a student for instance Name, address, class, section. Destination The destination address is defined in the header of the packet. The destination address contains the IP address of the final destination of the packet for delivery. Source It is also defined in the network header. The source address identifies the sender of the data packet. If there is NAT (Network Address Translation) involved then it is not necessary that the source address is correct because it will send the IP address of the NATing devices. Length The datagram size is defined in this field. The length field includes header and data in bytes. The minimum length of a data gram should be 20 bytes. PAD PAD is basically a packet assembler / Dissembler. It may be used as a software application or as a hardware device For Asynchronous transmission it formats the packet headers and also break down the data stream in to individual packets. CRC Cyclic redundancy check is commonly used for detecting the errors during the transmission of data. The messages which are transmitted are separated into prearranged lengths and are divided by a fixed divisor. The remainder number is sent along with the message as per calculation. When the message received by the computer, it re calculates the remainder and compares with the transmitter or source remainder, if the numbers not equivalent error occurs. b) What is the encoding scheme used with Cat 5, 100Base-TX? Solution: In Cat 5, 100Base-TX NRZI encoding is used on a physical medium attachment layer. One more encoding technique is used known as MLT-3. It is a medium-dependent sub layer which implies before the final encoding of the transmission on a maximum” fundamental frequency" of 31.25 MHz. c) Both nodes A and B attempt to transmit at time t=0 to the station C. After the collision, the nodes A and B draw from the back off interval the values of K=1 and K=0 respectively. At what time (in seconds) is the A’s frame completely delivered to C? You need to support your answer with a time space diagram. Solution: At time t = 0, both A and B transmit. At time t = 12.5 u Sec, A detects a Collision At time t = 25 u sec last bit of B’s aborted transmission arrives at A At time t = 37.5 u sec first bit of B’s aborted retransmission arrives at B At time t = 37.5u sec + (10,000 bits / 10x10x10x10x10x10x10 Bps) = 0.00100375 U sec d) If C acknowledges A, at what time will the acknowledgement arrive to A, from the time A has finished its transmission? Show this transmission on the same diagram of question (c). Solution: Question 2 The 1982 Ethernet specification, 10base, allowed between any two stations up to 1500m of coaxial cable, 1000 m of other point-to-point long cable, and two repeaters. Each station or repeater connects to the coaxial cable via up to 50 m of “drop cable”. Typical delays associated witch each device are 0.77c for the coaxial cable, 0.65c for the link/drop cable, 0.6sec each for the repeaters, and 0.2sec each for the transceivers. The constant c refers to the speed of the light in a vacuum = 3x108m/s. a) Including the 48-bits of jam signal, show that the total round-trip delays are closer to the official one of 52.1sec. Explain therefore why the minimum official packet size is set to 512bits. Solution: The transmitting node must be capable of transmitting the minimum packet size and reliably detecting whether a collision has occurred. The minimum transmission time necessary to transmit the minimum packet size is known as the slot time. For 10 / 100 Mbps the slot time is 512 bit times, which not coincidentally happens to be the minimum length of an Ethernet packet (64 octets is 512 bits). For gigabit Ethernet, the slot time is 4096 bit times. If a collision occurs, lot of things happen. If a transmitting node detects a collision, it transmits a jamming signal that is 32 bit times in length. The node stops transmitting and waits a random amount of times for the slot before trying to transmit again. This process is called back-off. Every collision that occurs while trying to transmit the same packet will trigger an increase in the back-off delay that the node may wait to try transmitting again. Back-off delay is a binary exponent range limited by the back-off limit, which is 10. For the first collision the node will wait for 0 or 1 slot times to retransmit again. After second collision, it will wait 0 to 3 slot times, in this way up to a maximum possible 1,023 slot times. If the frame cannot be retransmitted after 16 attempts, it is discarded and it will need to be handled by the upper layers in the OSI stack. b) What happen to the minimum packet size, if the delay time is held constant, and the transmission rate rises to 100Mbps? How do you achieve such a minimum packet size in practise? What is the drawback for such approach? Solution: If we use 48 bit of jam signal the minimum packet size would be 4640+ 48 bits = 586 Bytes. By this statement we can suppose that the sender will receive the whole jam signal frame to detect collision. The minimum packet size would be smaller if the length of the LAN is compact which therefore reduces the round trip propagation delay. The minimum packet size will result in waste of bandwidth. c) Recall that one way to achieve a transmission rate of 100 Mbps, using 100Base-T4, would be to use 4 set of category 3 unshielded twisted cables, operating in half duplex. Explain how each pair is used, and deduce how the rate 100Mbps can be achieved. Justify your answer, by providing some calculations. Solution: It operated over 4 pairs of 3, 4 or 5 UTP wiring. From these four pairs, three pairs are used for data transmission and one pair is used for collision detection. It is also capable to run on category 3 in this was it provides the easier migration to 100 BASE-T without re wiring and hence reducing time and cost. 3 pairs are engaged in the transmission of data so full duplex mode is not possible. The coding which is used is 8B6T. EIA/TIA-T568A PIN Signal Wire Colour 1 Transmit 3 White / green 2 Receive 3 green / White 3 Transmit 2 White / Orange 4 Receive 1 Blue / white 5 Transmit 1 White / Blue 6 Receive 2 Orange / White 7 Transmit 4 White / Brown 8 Receive 4 Brown / White 8B6T Encoding Technique: It is based on ternary symbol. It can take three values 1,0,-1 also represented as +, - , 0. d) Now suppose that the transmission rate rises to 1Gbps. How would this affect the minimum packet size? Is it normal to reduce the cable length to 100m? Justify your answer. Explain how such high rate can be achieved with 4 pairs of category 5 UTP, 1000Base-T. Solution: The 1000Base t physical layer standard provides 1 Gbps Ethernet signal transmission over the four pairs of Cat-5 unshielded twisted-pair (UTP) cable. The transmission on the network is at 125 M baud, it is the same symbol rate as 10 / 100 Fast Ethernet. By using more developed five-level pulse amplitude modulation (PAM-5) coding, it can transmit more data. To simplify, each wire pair sends and transmits simultaneously, for 250 Mbps per pair (125 M baud 2 bits). Multiplying 250 Mbps by four pairs yields the nominal rate of 1000 Mbps. Question 3 Many routers are now ‘MPLS capable’. Explain the difference between the way an MPLS capable router works and the way a normal router works. Why are MPLS capable routers replacing normal routers in many situations? Solution: What is MPLS? The Multiprotocol Label Switching (MPLS) protocol is a relatively new technology and it is a mechanism for labelling traffic flows to define a path through a complex network. A network operator can use MPLS as a management tool, as capacity and routes can be seen." (Shopforbandwidth, n.d.) How MPLS works? With frame relay, the provider's frame relay switches swap the frame packets based on data link connection identifier (DLCI) numbers. The switches don't look at the IP information, nor do they need to. With MPLS, this is not the case. The switches look at the IP source and destination of the packets because their IP routers are critical IP routers on your network. That means they must participate in whatever routing protocol we choose. If they're routing packets for us, they must have our routes. (Shopforbandwidth, n.d.) If we decide to migrate to an MPLS network provider, and we run Open Shortest Path First (OSPF) as our dynamic routing protocol, all sites have to have access to every other site. At the central site, we have an Internet circuit. (Shopforbandwidth, n.d.) Think of the MPLS provider's router as the "big" router in the cloud. All of our routers connect to its router, and we're routing the traffic together. Therefore, the MPLS providers are responsible for the routing. (Shopforbandwidth, n.d.) If we make changes to the routing protocol, we may need to request that the provider reconfigure its routing protocol. In most cases, however, we can just add a new LAN subnet without informing them at all, and the dynamic routing protocol does its job. Or, if we want to do this without involving the carrier, we could distribute a default route through the dynamic routing protocol and assuming the carrier doesn't have any filters, the remote routers should be able to access the Internet using this default route. (Shopforbandwidth, n.d.) Another option is to use the static route mentioned above. However, the next-hop router would be the carrier's router in the cloud, and the provider would have to configure a static route to point to our central router (to complete the connection). (Shopforbandwidth, n.d.) When working with MPLS, it's critical to remember that both our organization and our carrier are responsible for the routing of MPLS traffic. Some may view this is a drawback, but it has worked well for many organizations small to large. (Shopforbandwidth, n.d.) How MPLS capable router can replace normal routers? Reason 1: When using frame relay, we need to configure Layer 2 on your router, ending up with a WAN network with a connection to a central cloud. Through that cloud, we have connections (permanent virtual circuits or PVCs) to whichever remote sites we've requested and are paying for. If we have a fully meshed network (where every site has a connection to every other site), we have to pay for every one of those PVCs, which ultimately becomes quite expensive. (Shopforbandwidth, n.d.) In case of MPLS, we don't have to worry about PVCs—nor pay for them. The pricing structure of a MPLS circuit is similar to an Internet circuit where we just pay for the access loop and a connection fee. Once connected, we have a fully meshed IP network between all remote sites which ends up saving a lot of money if we have various locations with all sites being able to communicate with each other. (Shopforbandwidth, n.d.) However, it must be noted that MPLS circuits are not Internet circuits. While these circuits usually ride the same backbone as the carrier's Internet traffic, they aren't public circuits as they have a private IP address structure. Different network providers use MPLS in different ways. For example, when using AT&T, Verizon, Sprint, Qwest, Paetec, Savvis or any other MPLS providers, we actually run the MPLS protocol on the router and label (i.e., tag) every packet that goes out to the network. (Shopforbandwidth, n.d.) Reason 2: Another great benefit of MPLS is that we can use the full port speed of our circuits. There are no more committed information rates (CIRs) and no more forward explicit congestion notifications (FECNs) or backward explicit congestion notification (BECNs). Just like Internet service, if we buy a 1544-K T1 Internet circuit, we can use 1544 K of bandwidth, both upload and download. (Shopforbandwidth, n.d.) After setting up the MPLS connections, we have a fully meshed network between all remote sites. To route between all of these sites, we can use either static routes or a dynamic routing protocol such as EIGRP, OSPF, or BGP. With a dynamic routing protocol in place, every site can learn about the LAN's at every other site. With static routes, it is not the ideal option when you have more than a few sites especially if you want them to be fully meshed together. (Shopforbandwidth, n.d.) References Shopforbandwidth n.d., Differences between MPLS and frame relay, viewed 5 June 2009, http://www.shopforbandwidth.com/Differences-between-MPLS-and-frame-relay.php Read More